(a)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(b)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(c)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(d)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(e)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(f)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(g)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(h)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
Trending nowThis is a popular solution!
Chapter 3 Solutions
Organic Chemistry
- Show work.....don't give Ai generated solutionarrow_forwardShow work. Don't give Ai generated solutionarrow_forward10. Complete the following halogenation reactions for alkanes. Draw the structures of one of the many possible products for each reaction. Name the reactant and product. a) CH₂- CH-CH2-CH3 + Br₂ CH₂ UV UV b) + Cl2 c) CH3-CH₂ CHICHCHICH-CH CH₂-CH₂ + F2 UVarrow_forward
- Which of the following processes involves the largest photon energy? Group of answer choices Electron promotion from n=2 to n=5 Electron relaxing from n=4 to n=3 Ionization of an electron from n=2 Ionization of an electron from n=4arrow_forwardWhich of the following compounds does not match atomic ratio expectations in Mendeleev's 1872 periodic table? Group of answer choices NO2 Al2O3 SO3 CaOarrow_forwardNeed help with 14 and 15. 14. bromobenzene + (CHs),CuLi + THF / -78° followed by water quench is a. toluene else!! b. xylene c. cumene d. styrene e. something 15. When cumene + H,SO, / Na,Cr, 0,/water are mixed (refluxed) what is produced? a. 2-phenylpropanol phenol e. styrene b. benzoic acid c. no reaction!arrow_forward
- Which of the following orbitals intersect or overlap the x-axis in the standard cartesian coordinate system used? (Select ALL correct answers.) Group of answer choices px dxz dx2-y2 py dxy sarrow_forwardWhich of the following sets of elements is not a Dobereiner triad? (Choose the best answer.) Group of answer choices Li-Na-K Al-Ga-In Cr-Mo-W K-Rb-Csarrow_forwardDon't used Ai solution and don't used hand raitingarrow_forward
- Don't used hand raiting and don't used Ai solutionarrow_forwardGive the structure(s) of the product(s) the reaction below, and be sure to indicate any relative stereochemistry (you can assume that each of the Diels-Alder reactions will proceed with endo selectivity). Draw out relevant enantiomer(s) if they are expected to form. If no reaction is expected to occur under the indicated conditions, then write "no reaction" or NR, and explain why you would expect nothing to occur. If more than one product is formed, please indicate which one will be the major product or if they will be formed in equal amounts. In all cases, equimolar amounts of both components/reagents are present unless indicated otherwise I'm struggling to see how this reaction will go! I am wondering if it will cycle on itself but I'm not sure how I drew out a decagon but I'm a bit lostarrow_forwardGive the structure(s) of the product(s) for the reactions below, and be sure to indicate any relative stereochemistry (you can assume that each of the Diels-Alder reactions will proceed with endo selectivity). Draw out relevant enantiomer(s) if they are expected to form. If no reaction is expected to occur under the indicated conditions, then write "no reaction" or NR, and explain why you would expect nothing to occur. If more than one product is formed, please indicate which one will be the major product or if they will be formed in equal amounts. In all cases, equimolar amounts of both components/reagents are present unless indicated otherwise .arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY